How to find term of AP when other two terms of AP are given?

CBSE NCERT Solutions Chapter 5 Arithmetic Progressions Exercise 5.2 Question 8

8.   An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.

Solution:

An AP consists of 50 terms and the 50th term is equal to 106.

It is also given that $a_{3}=12$.           {$a_3$ is the 3rd term of AP}

Using formula ${{a}_{n}}=a+(n-1)d$,   to find nth term of arithmetic progression, we get

$a_{50} = a + (50-1)d$      And   $a_3 =a + (3-1)d$

$\Rightarrow 106 = a + (50-1)d$     And   $12 = a + (3-1)d$

$\Rightarrow 106 = a + 49d$    And  $12 = a + 2d$

These are equations consisting of two variables. Lets solve them using substitution method.

Using equation $106 = a + 49d$, we get   $a = 106-49d$

Putting value of $a$ in the equation $12 = a + 2d$, we get

$12 = 106-49d+2d$

$\Rightarrow 47d = 94$

$\Rightarrow d = \frac{94}{47} =2$

Putting value of $d$ in the equation, $a = 106-49d$, we get

$a = 106 -49 (2) = 106 - 98 = 8$

Therefore, First term $= a = 8$ and Common difference $= d = 2$

To find 29th term, we use formula ${{a}_{n}}=a+(n-1)d$ which is used to find nth term of arithmetic progression, we get

${{a}_{29}}=8+(29-1)2 = 8 + 56 = 64$

Therefore, 29th term of AP is equal to $64$.

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